Solve 3(a-7)^2+4+(a+3)(4-3a)

$$3 ( a - 7 ) ^ { 2 } + 4 + ( a + 3 ) ( 4 - 3 a )$$

Evaluate

$$163-47a$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-7\right)^{2}.

$$3\left(a^{2}-14a+49\right)+4+\left(a+3\right)\left(4-3a\right)$$

Use the distributive property to multiply 3 by a^{2}-14a+49.

$$3a^{2}-42a+147+4+\left(a+3\right)\left(4-3a\right)$$

Add 147 and 4 to get 151.

$$3a^{2}-42a+151+\left(a+3\right)\left(4-3a\right)$$

Use the distributive property to multiply a+3 by 4-3a and combine like terms.

$$3a^{2}-42a+151-5a-3a^{2}+12$$

Combine -42a and -5a to get -47a.

$$3a^{2}-47a+151-3a^{2}+12$$

Combine 3a^{2} and -3a^{2} to get 0.

$$-47a+151+12$$

Add 151 and 12 to get 163.

$$-47a+163$$

Expand

$$163-47a$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-7\right)^{2}.

$$3\left(a^{2}-14a+49\right)+4+\left(a+3\right)\left(4-3a\right)$$

Use the distributive property to multiply 3 by a^{2}-14a+49.

$$3a^{2}-42a+147+4+\left(a+3\right)\left(4-3a\right)$$

Add 147 and 4 to get 151.

$$3a^{2}-42a+151+\left(a+3\right)\left(4-3a\right)$$

Use the distributive property to multiply a+3 by 4-3a and combine like terms.

$$3a^{2}-42a+151-5a-3a^{2}+12$$

Combine -42a and -5a to get -47a.

$$3a^{2}-47a+151-3a^{2}+12$$

Combine 3a^{2} and -3a^{2} to get 0.

$$-47a+151+12$$

Add 151 and 12 to get 163.

$$-47a+163$$